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UPSC 2017 Maths Optional Paper 2 Q1b — Step-by-Step Solution

10 marks · Section A

Cayley's Theorem · Algebra · Read the full method →

Question

Let $G$ be a group of order $n$. Show that $G$ is isomorphic to a subgroup of the permutation group $S_n$.

Technique

Left-regular representation — $g\mapsto L_g$ (left multiplication); show it is a permutation, a homomorphism, with trivial kernel; conclude $G\hookrightarrow S_n$.

Solution

This is Cayley’s theorem. The idea: each $g\in G$ acts on the set $G$ by left multiplication, and this action is a permutation of the $n$-element set $G$; the map sending $g$ to that permutation is an injective homomorphism into $\operatorname{Sym}(G)\cong S_n$.

Step 1 — Left translation is a permutation of $G$

For $g\in G$ define $L_g:G\to G$ by $L_g(x)=gx$. Then $L_g$ is a bijection:

• Injective: $L_g(x)=L_g(y)\Rightarrow gx=gy\Rightarrow x=y$ (cancel $g$).
• Surjective: for any $y\in G$, $L_g(g^{-1}y)=g(g^{-1}y)=y$.

Hence $L_g\in\operatorname{Sym}(G)$, the group of all bijections of the set $G$ under composition. Since $|G|=n$, $\operatorname{Sym}(G)\cong S_n$ (relabel the $n$ elements of $G$ as $1,\ldots,n$).

Step 2 — The map $\Phi:G\to\operatorname{Sym}(G)$ is a homomorphism

Define $\Phi(g)=L_g$. For all $g,h,x\in G$,

so $L_{gh}=L_g\circ L_h$, i.e. $\Phi(gh)=\Phi(g)\Phi(h)$. Thus $\Phi$ is a group homomorphism. (Associativity of $G$ is exactly what makes this work; note $L_e=\mathrm{id}$ since $ex=x$.)

Step 3 — $\Phi$ is injective

Compute the kernel. If $\Phi(g)=L_g=\mathrm{id}$, then $gx=x$ for all $x\in G$; taking $x=e$ gives $g=ge=e$. Hence

so $\Phi$ is injective. (Equivalently, $\Phi$ is a monomorphism.)

Step 4 — Conclude via the first isomorphism theorem

$\Phi:G\to\operatorname{Sym}(G)$ is an injective homomorphism, so $G\cong\Phi(G)$, and $\Phi(G)$ is a subgroup of $\operatorname{Sym}(G)\cong S_n$. Therefore

Answer